Question 310678
{{{ graph( 300, 300, -10, 10, -10, 10, 2x^3-5x+7) }}}
Looks like there is a root near {{{x=-2}}}.
f(x)={{{2x^3 - 5x + 7}}}
f'(x)={{{6x^2-5}}}
I used Newton's method to iterate,
x(n+1)=xn-f(xn)/f'(xn)
Starting with {{{x=-2}}}, the solution quickly converge to {{{x=-2.05098}}} or to two decimal places, {{{highlight_green(x=-2.05)}}}
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{{{x(x-2)=7}}}
{{{x^2-2x-7=0}}}
Use the quadratic formula,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-(-2) +- sqrt( 4-4*1*(-7) ))/(2*1) }}}
{{{x = (2 +- sqrt( 32 ))/(2) }}} 
{{{x = (2 +- 4sqrt( 2 ))/(2) }}}
{{{x = 1 +- 2sqrt( 2 ) }}}